Derivation of Heat and mass transfer equation through different shapes and bodies; continuity equation & momentum equation;

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Heat and mass transfer equation
(a)Three-Dimensional heat transfer equation analysis (Cartesian co-ordinates)
Assumptions
• The solid is homogeneous and isotropic
• The physical parameters of solid materials are constant
• Steady state conduction
• Thermal conductivity k is constant
Consider a homogenous medium within which there is no bulk motion and the temperature
distribution T(x,y,z) is expressed in Cartesian coordinates.
We first define the infinitesimally small control volume dx.dy.dz, as shown in Figure. Energy
quantities are yields at,
Qx+ Qy +Qz+ Qgen= Qx+dx + Qy+dy+ Qz+dz+
𝑑𝐸
𝑑𝑡
--- (1)
Let, Cp = Specific heat of material J/Kg.0 C)
qg = Energy generation rate per unit volume W/m3
ρ = Density of material, Kg/m3
T = temperature, 0K
t = time,
k= Thermal Conductivity W/mK
x,y,z= Coordinates, m
Within the medium there may be an energy source term associated with the rate of thermal
energy generation, this term is represented by,
Qgen= qgdxdydz ---(2)
Where qg is the rate at which energy is generated per unit volume (W/m3)
In addition, changes may occur in the amount of the internal thermal energy stored by the
material in the control volume. If the material is not experiencing a change in phase, latent
energy effects are not pertinent, and the energy storage term may be expressed as,
𝑑𝐸
𝑑𝑡
= ρ Cp dx dy dz
𝑑𝑇
𝑑𝑡
---(3)

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where ρCp
𝑑𝑇
𝑑𝑡
is the time rate change of the sensible (thermal) energy of the medium per
unit volume.
Qx =-k dydz
𝑑𝑇
𝑑𝑥
Qy =-k dxdz
𝑑𝑇
𝑑𝑦
Qz= -k dxdy
𝑑𝑇
𝑑𝑧
Qx+dx = Qx+
𝑑𝑄𝑥
𝑑𝑥
Qy+dy = Qy+
𝑑𝑄𝑦
𝑑𝑦
Qz+dz = Qz+
𝑑𝑄𝑧
𝑑𝑧
Qx+dx = Qx - k dxdydzd2T/dx2 Qy+dy = Qy - k dxdydz d2T/dy2 Qz+dz= Qz - kdxdydzd2T/dz2
Substituting in equation (1) we get,
qgdxdydz= kdxdydz d2
T/dx2
- kdxdydz d2
T/dy2
- kdxdydz d2
T/dz2
+ ρCpdxdydz
𝑑𝑇
𝑑𝑡
k d2
T/dx + k d2
T/dy2
+ k d2
T/dz2
+ qg = ρCp
𝑑𝑇
𝑑𝑡
d2
T/dx + d2
T/dy2
+ d2
T/dz2
+
𝑞𝑔
𝑘
=
𝑑𝑇
𝑑 α 𝑡
---(4)
Where the quantity α =
𝑘
ρCp
(m2 /s) is called thermal diffusivity of the material.

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Spherical coordinates
Special cases
Steady state one dimensional heat flow (no heat generation)
Steady state one dimensional heat flow in cylindrical coordinates (no heat
generation)
Steady state one dimensional heat flow in Spherical coordinates (no heat generation)
Steady state one dimensional heat flow (with heat generation)

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Two-dimensional steady state heat flow (without heat generation)
(b)Heat conduction through a slab
Assumptions:
• One dimensional steady state heat transfer
• No heat generation
• The solid is homogeneous and isotropic
• The physical parameters of solid materials are constant
• Steady state conduction
• Thermal conductivity
k is constant One-dimensional heat transfer equation is,
Double integrating the equation,
C1 and C2 are the two constants, two boundary conditions are needed to determine the
constants Boundary conditions are,
Applying boundary conditions,

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Differentiating both sides
By Fourier law,
L/Ak = R is called as thermal resistance of the slab for heat flow through an area A across a
temperature T1-T2
This concept analogous to electric resistance in Ohm’s law as shown in figure
(c)Heat transfer through hollow cylinder
Assumptions:
• One dimensional steady state heat transfer, in r direction only
• No heat generation
• The solid is homogeneous and isotropic
• The physical parameters of solid materials are constant
• Thermal conductivity k is constant
• Temperature within the cylinder does not vary with time
One dimensional heat conduction equation,

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Boundary conditions,
Solving simultaneous equations,
By Fourier law,
is called as thermal resistance of the cylinder for heat flow through
an area A across a temperature Ti – To

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(d)Heat transfer through Sphere
r1, r2, inner and outer radii
Ti, To, inner and outer surface temperature
L, Length of cylinder
Assumptions:
• One dimensional steady state heat transfer, in r direction only
• No heat generation
• The solid is homogeneous and isotropic
• The physical parameters of solid materials are constant
• Thermal conductivity k is constant
One dimensional heat conduction equation,
-Double Integration
Boundary conditions,

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Solving simultaneous equations,
By Fourier law,
is called as thermal resistance of the Sphere for heat flow through
an area A across a temperature Ti – To
Composite slab

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**end of statement** please go to next page**

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Continuity Equation
The continuity equation is defined as the product of cross-sectional area of the pipe and the
velocity of the fluid at any given point along the pipe is constant.
Continuity equation represents that the product of cross-sectional area of the pipe and the
fluid speed at any point along the pipe is always constant. This product is equal to the
volume flow per second or simply the flow rate. The continuity equation is given as:
R = A v = constant
Where,
• R is the volume flow rate
• A is the flow area
• v is the flow velocity
Following are the assumptions of continuity equation:
• The tube is having a single entry and single exit
• The fluid flowing in the tube is non-viscous
• The flow is incompressible
• The fluid flow is steady
Derivation (Bernoulli’s Principle):
Now, consider the fluid flows for a short interval of time in the tube. So, assume that short
interval of time as Δt. In this time, the fluid will cover a distance of Δx1 with a velocity v1 at
the lower end of the pipe.
At this time, the distance covered by the fluid will be:

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Δx1 = v1Δt
Now, at the lower end of the pipe, the volume of the fluid that will flow into the pipe will
be:
V = A1 Δx1 = A1 v1 Δt
It is known that mass (m) = Density (ρ) × Volume (V). So, the mass of the fluid in Δx1 region
will be:
Δm1= Density × Volume
Δm1 = ρ1A1v1Δt --- (1)
Now, the mass flux has to be calculated at the lower end. Mass flux is simply defined as the
mass of the fluid per unit time passing through any cross-sectional area. For the lower end
with cross-sectional area A1, mass flux will be:
Δm1/Δt = ρ1A1v1 --- (2)
Similarly, the mass flux at the upper end will be:
Δm2/Δt = ρ2A2v2 --- (3)
Here, v2 is the velocity of the fluid through the upper end of the pipe i.e.
through Δx2 , in Δt time and A2, is the cross-sectional area of the upper end.
In this, the density of the fluid between the lower end of the pipe and the upper end of the
pipe remains the same with time as the flow is steady. So, the mass flux at the lower end of
the pipe is equal to the mass flux at the upper end of the pipe
i.e. Equation 2 = Equation 3.
Thus,
ρ1A1v1 = ρ2A2v2 --- (4)
This can be written in a more general form as:
ρ A v = constant
The equation proves the law of conservation of mass in fluid dynamics. Also, if the fluid is
incompressible, the density will remain constant for steady flow. So, ρ1 =ρ2.
Thus, Equation 4 can be now written as:
A1 v1 = A2 v2
This equation can be written in general form as:
A v = constant
Now, if R is the volume flow rate, the above equation can be expressed as:

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R = A v = constant
Continuity Equation in Cylindrical Coordinates
Special Cases
Following is the continuity equation for incompressible flow
∂ρ∂t+1r∂rρu∂r+1r∂ρv∂θ+∂ρw∂z=0
as the density, ρ = constant and is independent of space and time, we get:
∇.v = 0
Following is the continuity equation in cylindrical coordinates:
𝜕
𝜕𝑥
(ρu) +
𝜕
𝜕𝑦
(ρv) +
𝜕
𝜕𝑧
(ρw) =0
**end of statement** please go to next page**

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Derivation of momentum of equation:
From newtons 2nd law:
Newton's Second Law of motion states that the rate of change of momentum of an object
is directly proportional to the applied unbalanced force in the direction of the force.
If a body is subjected to multiple forces at the same time, then the acceleration produced is
proportional to the vector sum (that is, the net force) of all the individual forces.
ie- F =
𝑑𝑃
𝑑𝑡
=
𝑑𝑚𝑣
𝑑𝑡
p= momentum;
F = m
𝑑𝑉
𝑑𝑡
= m*a V = velocity;
F = m*a
Where,
F is the force applied,
m is the mass of the body,
and a, the acceleration produced.
Newton’s Second Law - The net force equals the rate of change of momentum
F⃗ =
𝑑𝑉⃗ 𝑀
d𝑡
For a system:
F⃗ =
𝑑
𝑑𝑡
∫SYSV⃗ dm
For a control volume (via Reynolds Transport Theorem):
∑F⃗ CV=
𝑑
𝑑𝑡
∫CVV⃗ ρdV+∫CSV⃗ ρV⃗ ⋅n^ dA

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(LHS 1) Body forces - weight for an element δm:
δF⃗ =
𝑑
𝑑𝑡
V⃗ δm=δm
𝑑
𝑑𝑡
V⃗ =δm⋅a⃗ δF =
𝑑
𝑑𝑡
V→δm=δm
𝑑
𝑑𝑡
V→=δm⋅a→
δF⃗ b=δm⋅g⃗ =ρδxδyδz⋅g⃗
(LHS 2) Normal force and Tangential force
Subscript notation:
• 1st subscript refers to the direction of the normal vector
• 2nd subscript refers to the direction of the stress vector
Sign convention:
• Normal Stress is positive if it’s in the same direction as the outward normal
vector
• Shear Stress is positive if it’s in the same direction as the coordinate system
w.r.t the outward normal vector (i.e. the right hand rule applies - thumb is the
direction of the outward normal vector, the two fingers are the shear stresses)
Conservation of Momentum in x-direction:
δFsx=(∂σxx/∂x+∂τyx/∂y+∂τzx/∂z)δxδyδz
Conservation of Momentum in y-direction:
δFsy=(∂τxy/∂x+∂σyy/∂y+∂τzy/∂z)δxδyδz
Conservation of Momentum in z-direction:
δFsz=(∂τxz/∂x+∂τyz/∂y+∂σzz/∂z)δxδyδz

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ρgx+∂σxx/∂x+∂τyx/∂y+∂τzx/∂z=ρ(∂u/∂t+u∂u/∂x+v∂u/∂y+w∂u/∂z)
ρgy+∂τxy/∂x+∂σyy/∂y+∂τzy/∂z=ρ(∂v/∂t+u∂v/∂x+v∂v/∂y+w∂v/∂z)
ρgz+∂τxz/∂x+∂τyz/∂y+∂σzz/∂z=ρ(∂w/∂t+u∂w/∂x+v∂w/∂y+w∂w/∂z)
from these three equations + continuity = Four equations we can conclude that-
Inviscid flow - no shearing stresses
σxx=σyy=σzz=−p
Euler’s Equation in the x-direction
ρgx−
𝛛𝐩
𝛛𝐱
= ρ(
𝛛𝐮
𝛛𝐭
+u
𝛛𝐮
𝛛𝐱
+v
𝛛𝐮
𝛛𝐲
+w
𝛛𝐮
𝛛𝐳
)
Vector Notation:
ρg⃗ −∇p=ρ(
∂V⃗
∂t
+V⃗ (∇⋅V⃗ )
**end**